another problem from giancoli

[oahsen]

there is another problem that i couldn't solve. it is about circular motion ;

a thin circular horizontal hoop of mass m and radius R rotates at frequency f about a vertical axis through its center. determine the tension within the hoop. (there is a hint : consider a tiny section of the hoop)
and the answer is 2*pi*m*R*f*f but how???????

[OlderDan]

there is another problem that i couldn't solve. it is about circular motion ;

a thin circular horizontal hoop of mass m and radius R rotates at frequency f about a vertical axis through its center. determine the tension within the hoop. (there is a hint : consider a tiny section of the hoop)
and the answer is 2*pi*m*R*f*f but how???????
Hint: Find the total force acting on half the hoop. Where does that force come from?

[oahsen]

thanks for hint but i didnt understand again. could you give a bit more explaination. i have midterm on friday and still i couldnt solve some problems and that makes me very sadddd

[NateTG]

Can you draw a free body diagram?

[oahsen]

yes i can draw but there is only m*v*v/r in the diagram. maybe there can a tension but if so they will be perpendicular and how can i write a relationship between them?

[OlderDan]

thanks for hint but i didnt understand again. could you give a bit more explaination. i have midterm on friday and still i couldnt solve some problems and that makes me very sadddd
You know the force on each little piece of the loop having mass dm = (m/2πr)dl = (m/2πr)(rdθ) is centripetal (dl is a tiny length of the arc of the hoop, and π is pi in a not very good font), so it can be set equal to d(m)v^2/r = (dm)rω^2. If you look at half the hoop, the force is in a different direction for each little piece, but by symmetry the total force is directed parallel to the axis of symmetry of the semicircle, i.e., in the direction of the force acting on the midpoint of the semicircle. If you break all the forces into components parallel and perpendicular to this direction, the sum of the perpendicular components is zero (by symmetry or by direct integration). The sum of the components parallel to this direction is the total force acting on that half of the hoop. The integral to find this sum is fairly easy. This force is provided by the other half of the hoop at the two points of contact where the halves are connected. At each contact point, the force is the tension in the hoop.

[oahsen]

ok i understood. 2*pi*f*f*r= (dm/dtheta)*(v^2)/r
=m * a ( for a piece an it is equal
to the tension )
thanks a lot